Answer :
Sure! Let's tackle this problem step-by-step to find the area of the isosceles right triangle formed by points [tex]\( A(-8, 7) \)[/tex], [tex]\( B(-4, 5) \)[/tex], and [tex]\( C(-2, 9) \)[/tex]. Here’s a detailed solution:
### Step 1: Plot and Label Points
1. Plot point [tex]\( A(-8, 7) \)[/tex].
2. Plot point [tex]\( B(-4, 5) \)[/tex].
3. Plot point [tex]\( C(-2, 9) \)[/tex].
Connect these points to form the triangle ABC.
### Step 2: Find the lengths of the sides
Using the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Calculate the distances:
1. Distance [tex]\( AB \)[/tex]:
[tex]\[ AB = \sqrt{((-4) - (-8))^2 + (5 - 7)^2} \][/tex]
[tex]\[ AB = \sqrt{(4)^2 + (-2)^2} \][/tex]
[tex]\[ AB = \sqrt{16 + 4} \][/tex]
[tex]\[ AB = \sqrt{20} \][/tex]
[tex]\[ AB = 4.47213595499958 \][/tex]
2. Distance [tex]\( BC \)[/tex]:
[tex]\[ BC = \sqrt{((-2) - (-4))^2 + (9 - 5)^2} \][/tex]
[tex]\[ BC = \sqrt{(2)^2 + (4)^2} \][/tex]
[tex]\[ BC = \sqrt{4 + 16} \][/tex]
[tex]\[ BC = \sqrt{20} \][/tex]
[tex]\[ BC = 4.47213595499958 \][/tex]
3. Distance [tex]\( CA \)[/tex]:
[tex]\[ CA = \sqrt{((-2) - (-8))^2 + (9 - 7)^2} \][/tex]
[tex]\[ CA = \sqrt{(6)^2 + (2)^2} \][/tex]
[tex]\[ CA = \sqrt{36 + 4} \][/tex]
[tex]\[ CA = \sqrt{40} \][/tex]
[tex]\[ CA = 6.324555320336759 \][/tex]
### Step 3: Identify the isosceles right triangle's legs
By examining the side lengths, we see that [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are equal, thus indicating that these are the legs of the isosceles right triangle.
### Step 4: Calculate the Area of the Triangle
For an isosceles right triangle, the area can be found using:
[tex]\[ \text{Area} = \frac{1}{2} \times (\text{leg})^2 \][/tex]
Here, each leg (AB or BC) has a length of [tex]\( 4.47213595499958 \)[/tex].
[tex]\[\text{Area} = \frac{1}{2} \times (4.47213595499958)^2 \][/tex]
[tex]\[\text{Area} = \frac{1}{2} \times 20 \][/tex]
[tex]\[\text{Area} = 10 \][/tex]
Therefore, the area of the right isosceles triangle formed by the points [tex]\( A(-8, 7) \)[/tex], [tex]\( B(-4, 5) \)[/tex], and [tex]\( C(-2, 9) \)[/tex] is [tex]\( 10 \)[/tex] square units.
### Step 1: Plot and Label Points
1. Plot point [tex]\( A(-8, 7) \)[/tex].
2. Plot point [tex]\( B(-4, 5) \)[/tex].
3. Plot point [tex]\( C(-2, 9) \)[/tex].
Connect these points to form the triangle ABC.
### Step 2: Find the lengths of the sides
Using the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Calculate the distances:
1. Distance [tex]\( AB \)[/tex]:
[tex]\[ AB = \sqrt{((-4) - (-8))^2 + (5 - 7)^2} \][/tex]
[tex]\[ AB = \sqrt{(4)^2 + (-2)^2} \][/tex]
[tex]\[ AB = \sqrt{16 + 4} \][/tex]
[tex]\[ AB = \sqrt{20} \][/tex]
[tex]\[ AB = 4.47213595499958 \][/tex]
2. Distance [tex]\( BC \)[/tex]:
[tex]\[ BC = \sqrt{((-2) - (-4))^2 + (9 - 5)^2} \][/tex]
[tex]\[ BC = \sqrt{(2)^2 + (4)^2} \][/tex]
[tex]\[ BC = \sqrt{4 + 16} \][/tex]
[tex]\[ BC = \sqrt{20} \][/tex]
[tex]\[ BC = 4.47213595499958 \][/tex]
3. Distance [tex]\( CA \)[/tex]:
[tex]\[ CA = \sqrt{((-2) - (-8))^2 + (9 - 7)^2} \][/tex]
[tex]\[ CA = \sqrt{(6)^2 + (2)^2} \][/tex]
[tex]\[ CA = \sqrt{36 + 4} \][/tex]
[tex]\[ CA = \sqrt{40} \][/tex]
[tex]\[ CA = 6.324555320336759 \][/tex]
### Step 3: Identify the isosceles right triangle's legs
By examining the side lengths, we see that [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are equal, thus indicating that these are the legs of the isosceles right triangle.
### Step 4: Calculate the Area of the Triangle
For an isosceles right triangle, the area can be found using:
[tex]\[ \text{Area} = \frac{1}{2} \times (\text{leg})^2 \][/tex]
Here, each leg (AB or BC) has a length of [tex]\( 4.47213595499958 \)[/tex].
[tex]\[\text{Area} = \frac{1}{2} \times (4.47213595499958)^2 \][/tex]
[tex]\[\text{Area} = \frac{1}{2} \times 20 \][/tex]
[tex]\[\text{Area} = 10 \][/tex]
Therefore, the area of the right isosceles triangle formed by the points [tex]\( A(-8, 7) \)[/tex], [tex]\( B(-4, 5) \)[/tex], and [tex]\( C(-2, 9) \)[/tex] is [tex]\( 10 \)[/tex] square units.
